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defon
3 years ago
11

Four students wrote different analogies to describe an electron before the formation of an ionic bond . Student A: A tug of war

between two wrestlers each weighing 460 lbs Student B: A dog and his owner tugging at a frisbee Student A big greedy dog stealing away a bone from a smaller dog Student D: Children exchanging their orange and strawberry flavored popsicles Which student presented a correct analogy ?
Chemistry
1 answer:
Nutka1998 [239]3 years ago
3 0

The correct analogy is that of "A big greedy dog stealing away a bone from a smaller dog."

An ionic bond is a bond formed formed between a metallic atom and a non-metallic atom.

The metallic atom is electropositive while the non-metallic atom is electronegative. Therefore, the metallic atom donates or gives up its electrons to the non-metallic atom which accepts the electrons.

During the formation of chemical bonds, only valence or outermost shell electrons are involved.

Metallic atoms have few valence electrons (between 1 to 3 electrons) while non-metallic atoms have many valence electrons (between 5 to 7 electrons).

Therefore the analogy by the student of a big greedy dog stealing away a bone from a smaller dog is correct because the non-metallic atoms takes the few electrons of the metallic atom and add to the many electrons they already have during the formation of an ionic bond.

Learn more about an ionic bond at: brainly.com/question/1225796

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Joy mixes one cup of sugar and one cup of lemon juice into three cups of water. The solvent in this recipe is the?
kobusy [5.1K]

Answer: Water is the solvent in this recipe.

Explanation: A solvent is " a molecule that has the ability to dissolve other molecules". Lemon juice and sugar are solutes.

5 0
3 years ago
How many atoms of each element are represented in Sn(CO3)2
Marina86 [1]
Sn=2
C and O=6
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3 0
3 years ago
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If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac
mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) +  6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0
4 years ago
Read 2 more answers
Plz help, I will give brainliest.
baherus [9]

Answer: The value of K_c is 2

Explanation:

Moles of  N_2O_4 = 1.0 mole  

Volume of solution = 1.00 L

Initial concentration of N_2O_4 = \frac{1.0mol}{1.00L}=1.0M

Equilibrium concentration of NO_2 = \frac{1.0mol}{1.00L}=1.0M  

The given balanced equilibrium reaction is,

                            N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initial conc.          1.0 M            0 M

At eqm. conc.     (1.0-x) M      (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Given : 2x = 1.0

x= 0.5

Now put all the given values in this expression, we get :

K_c=\frac{(1.0)^2}{(1.0-0.5)}

K_c=2

Thus the value of K_c is 2

4 0
3 years ago
What is the oxidation number of Pt in K₂PtCl₆?
Agata [3.3K]

Answer:

the oxidation number of Pt in K₂PtCl₆ <u>is 4.</u>

K₂PtCl₆=0

2+x-6=0

x=4

8 0
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